/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 A VW Beetle goes from 0 to \(60.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{~m} / \mathrm{s}^{2}\). (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{~s}\). Find the acceleration \(\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right)\) of the dragster.

Short Answer

Expert verified
(a) It takes 11.41 seconds for the Beetle. (b) The dragster's acceleration is 44.70 m/s².

Step by step solution

01

Convert Initial and Final Velocity Units

The Beetle's speed is given in miles per hour (mi/h). Convert this to meters per second (m/s) using the conversion factor: 1 mile = 1609.34 meters 1 hour = 3600 seconds. Thus, \(60.0 \text{ mi/h} = 60.0 \times \frac{1609.34}{3600} \approx 26.82 \text{ m/s} \).
02

Use the Formula for Time

Use the formula for acceleration, which is: \[ a = \frac{{v_f - v_i}}{t} \]where \( v_f = 26.82 \text{ m/s} \), \( v_i = 0 \text{ m/s} \), and \( a = 2.35 \text{ m/s}^2 \).Rearrange the formula to solve for time (\( t \): \[ t = \frac{{v_f - v_i}}{a} = \frac{{26.82 - 0}}{2.35} \approx 11.41 \text{ seconds} \].
03

Calculate the Dragster's Acceleration

For the dragster, we need to calculate the acceleration given time (\( t = 0.600 \text{ s} \)) and the same speed conversion for final velocity (\( v_f = 26.82 \text{ m/s} \)). The initial velocity is again 0.Use the formula: \[ a = \frac{{v_f - v_i}}{t} = \frac{{26.82 - 0}}{0.600} = 44.70 \text{ m/s}^2 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration
Acceleration is a fundamental concept in kinematics, which describes the change in velocity of an object over time. When you press the gas pedal in a car, the vehicle accelerates. This means there is an increase in speed. Acceleration can be both positive and negative, where positive acceleration means speeding up, and negative means slowing down.

The formula used to calculate acceleration is:
  • \( a = \frac{{v_f - v_i}}{t} \)
where:
  • \( a \) is the acceleration,
  • \( v_f \) is the final velocity,
  • \( v_i \) is the initial velocity,
  • \( t \) is the time period over which the change occurs.
In our example with the VW Beetle, we know the car's initial velocity \( v_i \) is 0 m/s and the final velocity \( v_f \) is calculated to be 26.82 m/s. The car accelerates at \( 2.35 \, \text{m/s}^2 \). You can see how the formula explains that the faster the velocity changes, the greater the acceleration.
Velocity Conversion
To analyze any problem in kinematics correctly, sometimes you need to convert units, especially when different units of measurements are used. Many times, speed is given in miles per hour (mi/h), but calculations often require meters per second (m/s) instead. The conversion helps keep the calculations consistent and easier to understand.

Here, the conversion process is critical. To convert 60.0 mi/h to m/s, use the conversion factors:
  • 1 mile = 1609.34 meters
  • 1 hour = 3600 seconds
The conversion would look like this:
  • \( 60.0 \, \text{mi/h} \times \frac{1609.34}{3600} = 26.82 \, \text{m/s} \)
Hence, converting velocity from mi/h to m/s ensures that all measures are in the same units, simplifying further calculations. Without converting, the application of the acceleration formula would be incorrect.
Calculating Time from Acceleration
The calculation of time when acceleration is known is handy. It allows us to determine how quickly an object gets to a certain velocity. In the scenario involving our VW Beetle, we use the equation for acceleration to solve for time. Rearranging the formula \( a = \frac{{v_f - v_i}}{t} \) to solve for \( t \) gives:
  • \( t = \frac{{v_f - v_i}}{a} \)

Please note:
  • \( v_f = 26.82 \, \text{m/s} \)
  • \( v_i = 0 \, \text{m/s} \)
  • \( a = 2.35 \, \text{m/s}^2 \)
By applying these values to the formula, the time \( t \) taken for the VW Beetle to reach its target speed is calculated as follows:
  • \( t = \frac{26.82}{2.35} \approx 11.41 \, \text{seconds} \)
Understanding how to manipulate this equation let us predict how long it takes to accelerate to a new speed. This application is valuable not just in physics education but also in practical, real-world scenarios autonomously, such as in sports or automotive testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) In general, does the average acceleration of an object have the same direction as its initial velocity \(v_{0}\), its final velocity \(v,\) or the difference \(v-v_{0}\) between its final and initial velocities? Provide a reason for your answer. (b) The following table lists four pairs of initial and final velocities for a boat traveling along the \(x\) axis. Use the concept of acceleration presented in Section 2.3 to determine the direction (positive or negative) of the average acceleration for each pair of velocities. $$ \begin{array}{|c|c|c|} \hline & \text { Initial velocity } v_{0} & \text { Final velocity } v \\ \hline(\mathrm{a}) & +2.0 \mathrm{~m} / \mathrm{s} & +5.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{b}) & +5.0 \mathrm{~m} / \mathrm{s} & +2.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{c}) & -6.0 \mathrm{~m} / \mathrm{s} & -3.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{d}) & +4.0 \mathrm{~m} / \mathrm{s} & -4.0 \mathrm{~m} / \mathrm{s} \\ \hline \end{array} $$ Problem The elapsed time for each of the four pairs of velocities is \(2.0 \mathrm{~s}\). Find the average acceleration (magnitude and direction) for each of the four pairs. Be sure that your directions agree with those found in the Concept Questions.

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The stones are thrown with the same speed. (a) Does the stone thrown upward gain or lose speed as it moves upward? Why? (b) Does the stone thrown downward gain or lose speed as time passes? Explain. (c) The speed at which the stones are thrown is such that they cross paths. Where do they cross paths, above, at, or below the point that corresponds to half the height of the cliff? Justify your answer. Problem The height of the cliff is \(6.00 \mathrm{~m},\) and the speed with which the stones are thrown is \(9.00 \mathrm{~m} / \mathrm{s}\). Find the location of the crossing point. Check to see that your answer is consistent with your answers to the Concept Questions.

You are on a train that is traveling at \(3.0 \mathrm{~m} / \mathrm{s}\) along a level straight track. Very near and parallel to the track is a wall that slopes upward at a \(12^{\circ}\) angle with the horizontal. As you face the window \((0.90 \mathrm{~m}\) high, \(2.0 \mathrm{~m}\) wide ) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner \(\mathrm{A}\) and eventually disappears at window corner \(\mathrm{B}\). How much time passes between appearance and disappearance of the upper edge of the wall?

In preparation for this problem, review Conceptual Example 7 . From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is \(9.0 \mathrm{~m} / \mathrm{s}\). (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain. (b) After \(0.50 \mathrm{~s}\), how far beneath the cliff top is the pebble?

From her bedroom window a girl drops a water-filled balloon to the ground, \(6.0 \mathrm{~m}\) below. If the balloon is released from rest, how long is it in the air?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.